The U.S. Public Health Service recommends the fluoridation of water as a means for preventing tooth decay. The recommended concentration is \(1 \mathrm{mg} \mathrm{F}^{-}\) per liter. The presence of calcium ions in hard water can precipitate the added fluoride. What is the maximum molarity of calcium ions in hard water if the fluoride concentration is at the USPHS recommended level? \(\left(K_{6}\right.\) for \(\mathrm{CaF}_{2}=4.0 \times 10^{-11}\) )

From the problem, we know that calcium ions react with fluoride ions to form insoluble calcium fluoride (CaF₂): \[ \mathrm{CaF}_{2}(s) \rightleftharpoons \mathrm{Ca}^{2+}(aq) + 2\mathrm{F}^{-}(aq) \] In order for this reaction to occur, hard water must already have some concentration of calcium ions, and we're going to find the maximum molarity of calcium ions without causing precipitation of CaF₂ in water.

The solubility product constant (Kₛₚ) for a compound can be written by multiplying the equilibrium concentrations of its dissociated ions raised to the power of their respective coefficients in the balanced equation. For CaF₂, the expression for the solubility product constant (Kₛₚ) will be: \[ K_{sp} = [\mathrm{Ca}^{2+}][\mathrm{F}^{-}]^2 \] The given Kₛₚ value for CaF₂ is \(4.0 \times 10^{-11}\).

From the problem, we know that the concentration of fluoride ions (F⁻) in the water is 1 mg/L, which is equivalent to 0.001 g/L. We need to convert this mass concentration to molarity. The molar mass of F⁻ is 19.00 g/mol, so: Fluoride concentration, \( [\mathrm{F}^{-}] =\frac{0.001 \ \mathsf{g/L}}{19.00 \ \text{g/mol}} = 5.26 \times 10^{-5} \ \text{M}\) Since the balanced equation shows that there are two moles of fluoride ions (F⁻) for every mole of calcium ions (Ca²⁺), the equilibrium concentration of calcium ions can be represented as x, while the equilibrium concentration of fluoride ions will be 2x. However, as there are already 5.26 × 10⁻⁵ M of fluoride ions present in the solution, the total fluoride ion concentration will be (2x + 5.26 × 10⁻⁵). Now, we can rewrite the solubility product constant (Kₛₚ) expression in terms of x: \[ K_{sp} = [\mathrm{Ca}^{2+}][(2x + 5.26 \times 10^{-5})]^2 \]

We will substitute the given Kₛₚ value (4.0 × 10⁻¹¹) into the expression and solve for x, the maximum molarity of calcium ions in the water without causing precipitation: \[ 4.0 \times 10^{-11} = [x][(2x + 5.26 \times 10^{-5})]^2 \] In this equation, x represents the equilibrium concentration of Ca²⁺ ions. It's reasonable to assume that the concentration of calcium ions will be the limiting factor in the precipitation of CaF₂, so we can approximate the equation by assuming x ≪ 5.26 × 10⁻⁵ M. The simplified equation will be: \[ 4.0 \times 10^{-11} = [x][ (5.26 \times 10^{-5})^2 ] \] Now, solve for x: \[ x = \frac{ 4.0 \times 10^{-11} }{ (5.26 \times 10^{-5})^2 } \approx 2.88 \times 10^{-2} \ \text{M} \]

The maximum molarity of calcium ions in hard water without causing precipitation of CaF₂ when the fluoride concentration is at the USPHS recommended level is approximately \(2.88 \times 10^{-2} \ \text{M}\).